# Amount of extension of the metal rods in the same system

If we heated two identical metal rod both vertically and horizontally, we would observe that it expands. However, if the rod is placed vertically, gravity will indeed play a role. The expansion of the rod increases its height, which in turn increases its gravitational potential energy. This is because the center of mass of the rod is raised higher. To clarify:

$$Q=mcΔT\ +\ ΔU$$

Now let's say we have two rectangular metal rods and their dimensions are L and L/4 as shown in the figure. The question I'm curious about is this: If we give a total heat Q to this system, what will be the amount of extension of these rods (Δh and ΔL)? (Assume that the heat is distributed properly and there is no energy loss in the system.And you don't need to give a value here, you can just use the units.)

The problem here is, I found Δh=ΔL. But I don't think this is correct. Eventually doesn't the metal block at the bottom do more work and need to extend less? Also if the temperature changes in the two blocks are different, the formula should be as follows

$$Q=mcΔT_{1}\ +mcΔT_{2}+\ ΔU$$

But can this formula be correct? After all, the two blocks are in contact and everything, including their initial temperatures, are identical and must be at the same temperature. So in short, what I mean is that it seems illogical to me in both cases.

• Why would the bottom block “need to extend less”? Commented Jan 17 at 16:53
• I don't know for sure, just an assumption. Problem is that the temperature change in both rods is equal, but as I mentioned in the question above, it does not gain any potential energy by extending, it is the extension of the lower rod that gives it this potential energy. So how do you make it evenly long? The lower rod both lengthens and gains potential energy, but as I said, the upper rod gains potential energy thanks to this rod, so we can not think of it as two separate systems. Commented Jan 17 at 18:27
• Are you assuming that the rods are perfectly rigid, or do they have a finite stiffness? Commented Jan 17 at 18:36
• yes they are perfectly rigid Commented Jan 17 at 18:39
• Great; that simplifies the solution to eliminate weight-driven strains. One more question: You didn't consider the thermal expansion in the rod width $L/4$. Would it be acceptable to assume that the width is very small relative to the length (so that thermal expansion of the width is negligible), or is it important for the rods to have a substantial width? Commented Jan 17 at 18:41

Assuming rigid rods, each of mass $$m$$, the constitutive law of strain development $$\varepsilon$$ in any direction from heating is $$\alpha\Delta T$$, where $$\alpha$$ is the thermal expansion coefficient and $$\Delta T$$ is the (uniform) temperature change.
You have correctly calculated that the energy increase $$\Delta U$$ from heating $$\Delta T$$ is $$2mc\Delta T+\frac{3}{2}\alpha mgL\Delta T=(2c+\frac{3}{2}\alpha gL)m\Delta T$$ (where $$c$$ is the specific heat capacity), comprising a sensible heat term and a potential energy term.
(If we wanted to acknowledge atmospheric pressure, we'd more precisely say that the enthalpy increase $$\Delta H$$ is $$(2c_P+\frac{3}{2}\alpha gL)m\Delta T$$, where $$c_P$$ is the constant-pressure specific heat capacity. This distinction is often ignored for condensed matter.)
This provides a new heat capacity for this particular system: $$(2c+\frac{3}{2}\alpha gL)m$$. You've shown that it takes more heating to obtain a given temperature increase, since some energy is necessarily diverted into potential energy, as mediated by the thermal expansion coefficient, the gravity field, and this particular geometry (e.g., the upper horizontal rod is lifted in toto, but only effectively half of the lower vertical rod rises this much, yielding the $$\frac{3}{2}$$ prefactor).